To find the equation of a line parallel to the line segment AB that passes through the point (1, 5), we first need to determine the slope of the line AB.
The slope \( m \) of a line through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For points A(6, 4) and B(9, 6):
- \( x_1 = 6, y_1 = 4 \)
- \( x_2 = 9, y_2 = 6 \)
Now, substituting these values into the slope formula:
\[ m = \frac{6 - 4}{9 - 6} = \frac{2}{3} \]
The slope of the line AB is \( \frac{2}{3} \).
Since we want the equation of a line parallel to AB, it will have the same slope \( \frac{2}{3} \).
Next, we can use the point-slope form of a line equation, which is given by:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1)\) is the point through which the line passes. Here, we will use point (1,5):
- \( m = \frac{2}{3} \)
- \( x_1 = 1, y_1 = 5 \)
Substituting these values into the point-slope form equation:
\[ y - 5 = \frac{2}{3}(x - 1) \]
Now let’s simplify this equation to get it into slope-intercept form \( y = mx + b \):
\[ y - 5 = \frac{2}{3}x - \frac{2}{3} \] \[ y = \frac{2}{3}x - \frac{2}{3} + 5 \] \[ y = \frac{2}{3}x - \frac{2}{3} + \frac{15}{3} \] \[ y = \frac{2}{3}x + \frac{13}{3} \]
Thus, the equation of the line parallel to line AB that passes through the point (1, 5) is:
\[ y = \frac{2}{3}x + \frac{13}{3} \]